# Can Linear Constant coefficient Difference Equations always represent an LTI system?

Can we say that a Linear Constant Coefficient Difference Equation can always represent a Linear Shift Invarient system ? Are there any conditions which need to be satisfied additionally by these kind of equations to be able to do that?

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http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/lecture-notes/MITRES_6_007S11_lec06.pdf

It is not necessary that a linear constant coefficient difference equation must represent an LTI system. It will represent an LTI system if and only if the solution satisfies the initial rest condition, namely if $x[n] = 0$ for $n<n_0$, then $y[n] = 0$ for $n<n_0$

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Just as a note, this initial rest condition is called causality. So, all causal LTI discrete time systems can be represented as a difference equation. –  Kartik Audhkhasi Aug 24 '12 at 20:55
@svenkatr Thnaks for the link. I guess the condition mentioned here represents a causal LTI system. Is it possible to represent an LTI system in general with the help of these equations (without imposing the additional constraint of causality) ? –  bubble Aug 25 '12 at 2:44
@bubble - causality is a necessary condition as well. –  svenkatr Aug 25 '12 at 19:11
@svenkatr Cant get you. Necessary for what ? –  bubble Aug 27 '12 at 2:48

WHat about final rest condition? (i.e if x(t) = 0, for t > t0 then y(t) is also 0 for t > t0 ) whether now the differential equation satisfies LSI or not?

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Welcome to MSE! In its current form, this is probably better as a comment as opposed to an answer (I realize you don't yet have enough reputation). perhaps you can improve it to make it an answer. Regards –  Amzoti May 23 '13 at 13:09
Thank you.Initial rest condition is not necessary condition for linearity. But it is essential for Time invariance to satisfy.For linearity alone, zero initial condition like y(1) = 0 (which is not initial rest condition) is enough. –  Ram May 25 '13 at 12:58